Professor, you have been an immense help this semester! Thank you kindly for your clear and concise explanations! Wishing you the best in all of your future endeavors.
Thank you a lot. The explanation is awesome, and I was finally able to understand how to work with Lemma. Though some difficulties I still have, for example the beginning exercise. Could anyone please explain why a the language comprised of words with equal number of 10's and 01's is regular? Thank you in advanced.
Hello from 7 years in the future. It has to do with how the only possible symbols we have are 0's and 1's. When you get a 01, the moment you get another 0 you'll also have a 10. In other words, a 010 contains both 01 and 10, so the only thing you have to do in your automat is wait until the correct letter comes and if it does, the number of 01 and 10's will be evened out. At least that's my interpretation.
Thank you But I think that we can Generate String with length < n and it contains a cycle(like self loops if we take it only two times and n sufficiently large then it will contains cycle), But we know for sure that |s| >= n iff it must be contains cycle
Thank you for the video. I have a question about your example which is to prove {0n1n} is not regular. You showed three cases that y might only consist of "1", or "0", or both "0" and "1". But according to the pumping lemma, the length of |xy|
+Yunhan Li I think he was just demonstrating that, if you were to ignore the third rule, then it still wouldn't work because the repeated y part would make the string unrecognisable by the machine. Of course, in an actual exam problem you wouldn't be able to split the string in such way because |xy|
+Matheus Gomes I recommend reading chapter 1.4 of "Introduction to Theory of Computation" by Sipser. There you can see many examples on how to use the pumping lemma for regular languages :)
But video by *hhp3* was uploaded in *2015.* The link you have mentioned has uploaded video in *2014.* It's difficult to *differentiate* between *original* and *repost.*
I think he used poor notation to describe this language, but since 01 and 10 configure a palindrome, they will belong to the Y part of your string, meaning that you can build a FSM with n states where Y loops through 1's AND Y's inbound state is 0 AND Y's outbound state is also 0. Ex.: (0)->(1*)->(0), outputs 011110. n(01) = n(10). I think this is it. Someone correct if I'm wrong.
Yeah I think that intuitively, any string that begins with 0 and ends with 0 will have an equal number of "01"s and "10"s. And the same is true for any string beginning with 1 and ending with 1. Making a FSM with those constraints is easier to think about.
Maybe this expression: (01)*0 can describe it? If you take for example 010 there's one occurence of 01 and one 10 or if you take 01010 there're two occurencies of both and so on. Even if (01)* returns empty string you end up with just 0 which means you have 0 occurences of both 01 and 10. I don't know if this way of thinking about this is correct though, but that's how I understand it. Another expression like (10)*1 can also work
Professor, you have been an immense help this semester! Thank you kindly for your clear and concise explanations! Wishing you the best in all of your future endeavors.
Perfect explanation of the pumping lemma. Thanks a lot!
Thank you so much, dear Professor! God bless you!
Thank you for the video - I had never seen Pumping Lemma explained this way before and now I understand
Just wanted to let you know that you saved me this semester !! Thank you for the videos
This video is so good! I was struggling with this. It was only after watching this that I finally got it!
thank you for the video, i really understood how use the lemma !!
Thanks a million. The explanation was very clear, Now I understood
This is a very helpful course. Thanks for your help professor!
A.I student from Utrecht University thanking u ! I'm going to recommend you to my teachers as an example.
Thank you, Harry.
I liked this explanation. So helpful
thank you ...perfect teaching ..saved my time and it was really goooood
Thank you so much. Great explanation!
This is very well explained. Thank you.
would love to see your tutorials on compiler theory,
Thank you, this is very useful
You're a wizard Harry! (harry porter, harry potter)
Thank you.
Thanks so much!
The part about "y" being interpreted as different parts of the string was a little bit confusing, because the cases 1, 2 and 3 did not seem to match.
thanks it was great
thank u alot
8:58 Why.. why!!?
I'm so grateful for these videos..Thank you very much.:)
amazing
Thank you a lot. The explanation is awesome, and I was finally able to understand how to work with Lemma. Though some difficulties I still have, for example the beginning exercise.
Could anyone please explain why a the language comprised of words with equal number of 10's and 01's is regular?
Thank you in advanced.
+Radu - Andrei Coandă because we can get FSA for that language, you can try in a paper to get one hwo recognise this language
Hello from 7 years in the future. It has to do with how the only possible symbols we have are 0's and 1's. When you get a 01, the moment you get another 0 you'll also have a 10. In other words, a 010 contains both 01 and 10, so the only thing you have to do in your automat is wait until the correct letter comes and if it does, the number of 01 and 10's will be evened out. At least that's my interpretation.
Hello. Do you have a video on Myhill Nerode's theroem?
there's a channel who has stolen your video
@@rj-nj3uk except they don't give credits.
A language with string length is 3 cannot be regular in that situtation, I cant get it
Can we use pumping lema to prove regularity?
It would seem that this professor is actually interested in making himself understood, and stands out for that reason alone.
Thank you
But I think that we can Generate String with length < n and it contains a cycle(like self loops if we take it only two times and n sufficiently large then it will contains cycle), But we know for sure that
|s| >= n iff it must be contains cycle
Thank you for the video. I have a question about your example which is to prove {0n1n} is not regular. You showed three cases that y might only consist of "1", or "0", or both "0" and "1". But according to the pumping lemma, the length of |xy|
+Yunhan Li I think he was just demonstrating that, if you were to ignore the third rule, then it still wouldn't work because the repeated y part would make the string unrecognisable by the machine. Of course, in an actual exam problem you wouldn't be able to split the string in such way because |xy|
+Matheus Gomes I recommend reading chapter 1.4 of "Introduction to Theory of Computation" by Sipser. There you can see many examples on how to use the pumping lemma for regular languages :)
Can someone please help me to find pumping length of 0*1+0+1*+10*1
Someone reposted your video here: th-cam.com/video/g4e2RElzCSQ/w-d-xo.html and received a lot more views off of it.
But video by *hhp3* was uploaded in *2015.* The link you have mentioned has uploaded video in *2014.*
It's difficult to *differentiate* between *original* and *repost.*
I cannot create a FSM which recognizes only strings with an equal number of 01s and 10s, without some sort of counting. Did anybody figure it out?
I think he used poor notation to describe this language, but since 01 and 10 configure a palindrome, they will belong to the Y part of your string, meaning that you can build a FSM with n states where Y loops through 1's AND Y's inbound state is 0 AND Y's outbound state is also 0. Ex.: (0)->(1*)->(0), outputs 011110. n(01) = n(10). I think this is it. Someone correct if I'm wrong.
Yeah I think that intuitively, any string that begins with 0 and ends with 0 will have an equal number of "01"s and "10"s. And the same is true for any string beginning with 1 and ending with 1. Making a FSM with those constraints is easier to think about.
Maybe this expression: (01)*0 can describe it? If you take for example 010 there's one occurence of 01 and one 10 or if you take 01010 there're two occurencies of both and so on. Even if (01)* returns empty string you end up with just 0 which means you have 0 occurences of both 01 and 10. I don't know if this way of thinking about this is correct though, but that's how I understand it. Another expression like (10)*1 can also work
Actually, now that I think about it, the better expression would be: ((01)*0) + ((10)*1) + (0*) + (1*)
Um... what is wrong with (0110)*?